32 Statics. 



F^g. 18, 53^ If from any point F taken in the plane of any parallelogram 

 A BCD, we let fall upon the contiguous sides AB, AD, and the 

 diagonal AC, the perpendiculars FE, FH, FG, the sum of the pro- 

 ducts of the two contiguous sides by the perpendiculars respectively let 

 fall upon them, will be equal to the product of the diagonal by its per- 



Fig. 18-pendicular, when the point F is neither in the angle BAD, nor in 

 the vertical angle KAL. If) on the contrary, the point F is either 

 ' in the angle BAD, or in the vertical angle KAL, the difference of 

 the products of the two contiguous sides by their respective perpendic- 

 ulars will be equal to the product of the diagonal by the perpendicu* 

 lar let fall upon it. 



Produce the side BC till it meets in / the perpendicular 

 and join FA, FB, FC, FD. The triangle 



Fig. is. FAC = FAB + ABC + FBC = FAB + ADC + FBC. 

 Now 



Geom: 1. The triangle FAC = 12L 



176. ;'" 



2. The triangle FAB = AB * FE . 



3. The triangle ADC having AD for its base, and IH for its 

 altitude, 



" 



4. The triangle F. BC = BC X FI 



o 



2 2 



Whence 



AC x FG _ AB x FE AD x IH AD x FI 



2 2 2 ~~2 " 



Now IH -}- FI =. FH; therefore, by doubling the whole, we 

 have 



: FH. 



Fig. 19. With respect to the triangle FAC, we have 



FAC = ABC F^J5 FBC = ADC F^B FBC, 

 that is, 



j3C x FG _ AD x IH ABx FE BC x FI 



~~2~~ 2 2 ~~2 ^ 



or, since J5C = ./ID, and IH FI = FF, the whole being 

 doubled, 



^C x FG = ADxFHABx FE. 



